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In a cubic system with cell edge a, two phonons with wave vectors $${\overrightarrow {\bf{q}} _1}$$ and $${\overrightarrow {\bf{q}} _2}$$ collide and produce a third phonon with a wave. vector $${\overrightarrow {\bf{q}} _3}$$ such that $${\overrightarrow {\bf{q}} _1} + {\overrightarrow {\bf{q}} _2} = {\overrightarrow {\bf{q}} _3} + \overrightarrow {\bf{R}} $$ where, $$\overrightarrow {\bf{R}} $$ is a lattice vector. Such a collision process will lead to(a)
A
finite thermal resistance
B
zero thermal resistance
C
an infinite thermal resistance
D
a finite thermal resistance for certain $$\left| {\overrightarrow {\bf{R}} } \right|$$ only
Correct Answer:
finite thermal resistance
The primitive translation vectors of the body centred cubic lattice are $$\overrightarrow {\bf{a}} = \frac{a}{2}\left( {{\bf{\hat x}} + {\bf{\hat y}} - {\bf{\hat z}}} \right),\,\overrightarrow {\bf{b}} = \frac{a}{2}\left( { - {\bf{\hat x}} + {\bf{\hat y}} + {\bf{\hat z}}} \right)$$ and $$\overrightarrow {\bf{c}} = \frac{a}{2}\left( {{\bf{\hat x}} - {\bf{\hat y}} + {\bf{\hat z}}} \right)$$ . The primitive translation vectors $$\overrightarrow {\bf{A}} ,\,\overrightarrow {\bf{B}} $$ and $$\overrightarrow {\bf{C}} $$ of the reciprocal lattice are
A
$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} + {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat z}}} \right)$$
B
$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} - {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat z}}} \right)$$
C
$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} + {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} - {\bf{\hat z}}} \right)$$
D
$$\overrightarrow {\bf{A}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat y}}} \right);\,\overrightarrow {\bf{B}} = \frac{{2\pi }}{a}\left( {{\bf{\hat y}} + {\bf{\hat z}}} \right);\,\overrightarrow {\bf{C}} = \frac{{2\pi }}{a}\left( {{\bf{\hat x}} + {\bf{\hat z}}} \right)$$
The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$ where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$ is the angular momentum. The Hamiltonian does not commute with
A
$$\overrightarrow {\bf{L}} + \overrightarrow {\bf{S}} $$
B
$$\overrightarrow {{{\bf{S}}^2}} $$
C
$${L_z}$$
D
$$\overrightarrow {{{\bf{L}}^2}} $$
If for a system of N particles of different masses m
1
, m
2
, . . . m
N
with position vectors $${\overrightarrow {\bf{r}} _1},\,{\overrightarrow {\bf{r}} _2},\,.\,.\,.\,{\overrightarrow {\bf{r}} _N}$$ and corresponding velocities $${\overrightarrow {\bf{v}} _1},\,{\overrightarrow {\bf{v}} _2},\,.\,.\,.\,{\overrightarrow {\bf{v}} _N}$$ respectively such that $$\sum\limits_i {\overrightarrow {{{\bf{v}}_i}} = 0,} $$ then
A
total momentum must be zero
B
total angular momentum must be independent of the choice of the origin
C
the total force on the system must be zero
D
the total torque on the system must be zero
How does a host on an Ethernet LAN know when to transmit after a collision has occurred?
1. In a CSMA/CD collision domain, multiple stations can successfully transmit data simultaneously.
2. In a CSMA/CD collision domain, stations must wait until the media is not in use before transmitting.
3. You can improve the CSMA/CD network by adding more hubs.
4. After a collision, the station that detected the collision has first priority to resend the lost data.
5. After a collision, all stations run a random backoff algorithm. When the backoff delay period has expired, all stations have equal priority to transmit data.
A
1 and 3
B
2 and 4
C
1,3 and 4
D
2 and 5
For two non-zero vectors $$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$, if $$\overrightarrow {\text{A}} $$ + $$\overrightarrow {\text{B}} $$ is perpendicular to $$\overrightarrow {\text{A}} $$ - $$\overrightarrow {\text{B}} $$ then,
A
Magnitude of $$\overrightarrow {\text{A}} $$ is twice magnitude of $$\overrightarrow {\text{B}} $$
B
Magnitude of $$\overrightarrow {\text{A}} $$ is half the magnitude of $$\overrightarrow {\text{B}} $$
C
$$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$ cannot be orthogonal
D
the magnitudes of $$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$ are equal
Due to the technical snag in the signal system two trains start approaching each other on the same track from two different stations, 240 km away each other. When the train starts a bird also starts moving to and fro between the two trains at 60 kmph touching each train each time. The bird initially sitting on the top of the engine of one of the trains and it moves so till these trains collide. If these trains collide one and half hour after start, then how many kilometers bird travels till the time of collision of trains?
A
90 km
B
130 km
C
120 km
D
95 km
E
None of these
Let $$\nabla \cdot \left( {{\text{f}}\overrightarrow {\text{v}} } \right) = {{\text{x}}^2}{\text{y}} + {{\text{y}}^2}{\text{z}} + {{\text{z}}^2}{\text{x}},$$ where f and v are scalar and vector fields respectively. If $$\overrightarrow {\text{v}} = {\text{y}}\overrightarrow {\text{i}} + {\text{z}}\overrightarrow {\text{j}} + {\text{x}}\overrightarrow {\text{k}} ,$$ then $$\overrightarrow {\text{v}} \cdot \nabla {\text{f}}$$ is
A
x<sup>2</sup>y + y<sup>2</sup>z + z<sup>2</sup>x
B
2xy + 2yz + 2zx
C
x + y + z
D
0
Consider two particles with position vectors $$\overrightarrow {{{\bf{r}}_1}} $$ and $$\overrightarrow {{{\bf{r}}_2}} $$ . The force exerted by particle 2 on particle 1 is $$\overrightarrow {\bf{F}} \left( {\overrightarrow {{{\bf{r}}_1}} ,\,\overrightarrow {{{\bf{r}}_2}} } \right) = \left( {{{{\bf{\dot r}}}_2} - {{{\bf{\dot r}}}_1}} \right)\left( {{r_2} - {r_1}} \right).$$ The force is
A
central and conservative
B
non-central and conservative
C
central and non-conservative
D
non-central and non-conservative
In a cubic crystal, atoms of mass M
1
lie on one set of planes and atoms of mass M
2
lie on planes interleaved between those of the first set. If C is the forte constant between nearest neighbour planes, the frequency of lattice vibrations for the optical phonon branch with wave vector k = 0 is
A
$$\sqrt {2C\left( {\frac{1}{{{M_1}}} + \frac{1}{{{M_2}}}} \right)} $$
B
$$\sqrt {C\left( {\frac{1}{{2{M_1}}} + \frac{1}{{{M_2}}}} \right)} $$
C
$$\sqrt {C\left( {\frac{1}{{{M_1}}} + \frac{1}{{2{M_2}}}} \right)} $$
D
zero
An atom with net magnetic moment $$\overrightarrow \mu $$ and net angular momentum $$\overrightarrow {\bf{L}} \left( {\overrightarrow \mu = \gamma \overrightarrow {\bf{L}} } \right)$$ is kept in a uniform magnetic induction $$\overrightarrow {\bf{B}} = {B_0}{\bf{\hat k}}.$$ The magnetic moment $$\overrightarrow \mu \left( { = {\mu _x}} \right)$$ is
A
$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + \gamma {B_0}{\mu _x} = 0$$
B
$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + \gamma {B_0}\frac{{d{\mu _x}}}{{dt}} + {\mu _x} = 0$$
C
$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + {\gamma ^2}B_0^2{\mu _x} = 0$$
D
$$\frac{{{d^2}{\mu _x}}}{{d{t^2}}} + 2\gamma {B_0}{\mu _x} = 0$$